证明∫x(f(x)^2)dx/∫xf(x)dx≤∫f(x)^2dx/∫f(x)dx(下线均为0.上限均为1)
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证明∫x(f(x)^2)dx/∫xf(x)dx≤∫f(x)^2dx/∫f(x)dx(下线均为0.上限均为1)
f(x)为在[0,1]上单调减少且恒大于零的连续函数
f(x)为在[0,1]上单调减少且恒大于零的连续函数
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由f(x) > 0,原式等价于(∫{{0,1} t·f(t)²dt)·(∫{0,1} f(t)dt) ≤ (∫{{0,1} f(t)²dt)·(∫{0,1} t·f(t)dt).
设F(x) = (∫{{0,x} f(t)²dt)·(∫{0,x} t·f(t)dt)-(∫{{0,x} t·f(t)²dt)·(∫{0,x} f(t)dt),则F(0) = 0.
F'(x) = f(x)²·(∫{0,x} t·f(t)dt)+x·f(x)·(∫{{0,x} f(t)²dt)-x·f(x)²·(∫{0,x} f(t)dt)-f(x)·(∫{{0,x} t·f(t)²dt)
= f(x)·∫{0,x} (t·f(x)+x·f(t)-x·f(x)-t·f(t))·f(t) dt
= f(x)·∫{0,x} (x-t)(f(t)-f(x))·f(t) dt.
对t∈[0,x],有x-t ≥ 0,而f(x)单调减,故f(t)-f(x) ≥ 0,又f(x) > 0,f(t) > 0.
于是有F'(x) = f(x)·∫{0,x} (x-t)(f(t)-f(x))·f(t) dt ≥ 0.
因此F(x)单调增,有F(1) ≥ F(0) = 0.
即(∫{{0,1} t·f(t)²dt)·(∫{0,1} f(t)dt) ≤ (∫{{0,1} f(t)²dt)·(∫{0,1} t·f(t)dt).
设F(x) = (∫{{0,x} f(t)²dt)·(∫{0,x} t·f(t)dt)-(∫{{0,x} t·f(t)²dt)·(∫{0,x} f(t)dt),则F(0) = 0.
F'(x) = f(x)²·(∫{0,x} t·f(t)dt)+x·f(x)·(∫{{0,x} f(t)²dt)-x·f(x)²·(∫{0,x} f(t)dt)-f(x)·(∫{{0,x} t·f(t)²dt)
= f(x)·∫{0,x} (t·f(x)+x·f(t)-x·f(x)-t·f(t))·f(t) dt
= f(x)·∫{0,x} (x-t)(f(t)-f(x))·f(t) dt.
对t∈[0,x],有x-t ≥ 0,而f(x)单调减,故f(t)-f(x) ≥ 0,又f(x) > 0,f(t) > 0.
于是有F'(x) = f(x)·∫{0,x} (x-t)(f(t)-f(x))·f(t) dt ≥ 0.
因此F(x)单调增,有F(1) ≥ F(0) = 0.
即(∫{{0,1} t·f(t)²dt)·(∫{0,1} f(t)dt) ≤ (∫{{0,1} f(t)²dt)·(∫{0,1} t·f(t)dt).
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